Expected value of occurrences of a variable under a Poisson distribution

Question

I asked a question on this forum recently, to which I got a very thorough answer that I didn't properly understand. Link to the original question here.

The answer stated:

"If we have N buckets and place M entries in them, the number of elements in each buckets, for large $N, m$, approximately follows ("Poissonization") a Poisson distribution $P_k= e^{-\lambda} \frac{\lambda^k}{k!}$ with $λ=M/N$.

If we denote $X_i=1$ if the bucket i is empty, 0 otherwise, we have $E[X_i]≈e^{\frac{−M}{N}}$"

How did they derive $E[X_i]≈e^{\frac{−M}{N}}$? I have tried and failed to deduce it myself or find online resources.

Answer 1

If you take the Poisson approximation then $$E[X_i] = P_0 \approx e^{-M/N} \frac{(M/N)^0}{0!}= e^{-M/N}$$

If you want the binomial route then $$E[X_i] = P(\text{no balls go in that bucket }) = \left(1-\frac1N\right)^M \approx e^{-M/N}$$ for large $M$ and $N$

Answer 2

The expected value in each bucket is $\frac MN$. Using the Poisson formula, the probability of getting $0$ in a bucket is $$ e^{-\frac MN}\frac{\left(\frac MN\right)^0}{0!}=e^{-\frac MN}\tag1 $$ A way to visualize this is to note that the probability that placing an entry misses our bucket is $\left(1-\frac1N\right)$. Thus, the probability that all of $M$ entries miss our bucket is $$ \left(1-\frac1N\right)^M\approx\left(e^{-\frac1N}\right)^M=e^{-\frac MN}\tag2 $$